2006-11-02 07:42:32 · 3 answers · asked by lilv 1 in Science & Mathematics ➔ Physics
Lead has the combined properties of being soft and dense. As a result, a large bulk of lead cannot support its own weight. If pure lead the size of a mountain were placed on the surface of the Earth, it would quickly yield in compression at the base due to the incredible pressure of the extremely massive quantity of lead above it. The base of the mountain would squash outwards, and this would happen irregularly and unpredictably, completely destabilizing the upper portion of the mountain, and the whole thing would collapse very spectacularly.
Lead has an ultimate tensile strength (UTS) of 19 MPa and a density of 11.34 g/cc = 11340 kg/m^3. Pressure is equal to rho*gh, where rho is density, g is gravity, and h is height of material. Solving P = UTS ==> rho*gh = UTS ==> (11340 kg/m^3)(9.8 m/s^2)h = 19000000 Pa ==> h = (19000000 Pa) / ((11340 kg/m^3)(9.8 m/s^2) = 171 m, the tallest a block of lead could be without the pressure at the base exceeding the material's strength. This is taller than the Great Pyramid of Giza, but not as tall as the Eiffel Tower, so would not qualify as a mountain by most definitions. This maximum height is inversely proportional to the acceleration due to gravity, so a taller lead mountain could exist somewhere other than Earth, where gravity was less powerful.
To compare, stainless steel has UTS = 620 MPa and rho = 8 g/cc, giving an allowable height of 7,908 m, which would actually make it the 18th tallest mountain in the world.
2006-11-08 02:31:45 · answer #1 · answered by DavidK93 7 · 0⤊ 0⤋
It's a heavy metal, hence you wouldn't have nearly enough of it to do so.
Where would you find such a lead deposit on the earths surface?
2006-11-02 16:05:01 · answer #2 · answered by Anonymous · 0⤊ 0⤋
because of its density and mass, it would sink into the land over time.
2006-11-02 15:44:20 · answer #3 · answered by papeche 5 · 0⤊ 0⤋